An equiconvex lens made of glass with a refractive index of 1.5 has a certain optical power. It is replaced by a new biconvex lens whose radii of curvature are R and R 2 , where R is the radius of curvature of the original equiconvex lens. If the power of the new lens is identical to that of the original lens, the refractive index of the material of the new lens is :

Step-by-step solution

Let the radius of curvature of each surface of the equiconvex lens be R . Using the lens maker's formula, the power of the first lens is: P₁ = ( ₁ - 1) ( 1 R₁ - 1 R₂ ) P₁ = (1.5 - 1) ( 1 R - 1 -R ) = 0.5 2 R = 1 R For the second biconvex lens, the radii of curvature are R₁ = R and R₂ = - R 2 . Its power is: P₂ = ( ₂ - 1) ( 1 R - 1 - R 2 ) = ( ₂ - 1) ( 1 R + 2 R ) = ( ₂ - 1) 3 R Since the powers are identical, P₁ = P₂ : 1 R = ( ₂ - 1) 3 R ₂ - 1 = 1 3 ₂ = 4 3 Answer: 4 3